Roller Coaster Physics: Force & Max Speed

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SUMMARY

The discussion focuses on calculating the forces acting on a roller coaster at two points, A and B, using the equation EF = m(v²/r). At point A, with a mass of 510 kg and a speed of 19.6 m/s, the force exerted by the track is calculated incorrectly as 19592.16 N. The correct approach requires accounting for gravitational acceleration and centripetal acceleration. For point B, the maximum speed must be determined by balancing the forces, specifically using the relationship v²/15 = g, where g is the acceleration due to gravity.

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Homework Statement


A roller coaster has a mass of 510kg when fully loaded with passengers. The radius at point A is 10.0 m and the radius at point B is 15.0 m.

-If the vehicle has a speed of 19.6 m/s at point A, what is the force exerted by the track on the car at this point? (N)

-What is the maximum speed the vehicle can have at B and still remain on the track? (m/s)



Homework Equations



EF=m (v^2 / r)

The Attempt at a Solution



*For the first problem I thought you would use the equation EF=m (v^2 / r). Which would be EF=510 (19.6^2 / 10.0). Which equals 510*(384.16/10.0). Equals 19592.16. However, this is wrong! Isn't that the right equation?

*For the second problem, I'm not sure where to start. But I'm assuming that I will need the correct force value before I can begin.

Any suggestion as to where I'm going wrong?
 
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It all depends on the location of points A and B, perhaps if you could post a diagram or describe the position of the two points we could help you further.
 
assuming a is on a curve that is convex up, (try it the other way and you'll find its off the track i think), you have two a accelerations to account for, g and the v^2/r which is 19.6^2/10=38.4. Compute the net a, then you have the force pushing on the track.

b must be convex down (or upside down, but then it would be a minimal speed): v^2/15=g as the two accelerations are opposing.
 

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